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(A, D) The total time of flight $T$ is given as $12 	ext{ seconds}$.For a particle projected vertically upwards,the time to reach the maximum height

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greatest height attained is $576 	ext{ ft}$

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(A, D) The total time of flight $T$ is given as $12 	ext{ seconds}$.For a particle projected vertically upwards,the time to reach the maximum height is $t = \frac{T}{2} = \frac{12}{2} = 6 	ext{ seconds}$.At maximum height,the final velocity $v = 0$.Using the equation $v = u - gt$,where $g = 32 	ext{ ft/sec}^2$:$0 = u - (32)(6) \Rightarrow u = 192 	ext{ ft/sec}$.Now,the maximum height $H$ is given by $H = ut - \frac{1}{2}gt^2$:$H = (192)(6) - \frac{1}{2}(32)(6)^2 = 1152 - 576 = 576 	ext{ ft}$.Thus,the velocity of projection is $192 	ext{ ft/sec}$ and the greatest height attained is $576 	ext{ ft}$.Both options $A$ and $D$ are correct.

$A$ particle is projected vertically upwards. If it has to stay above the ground for $12 \text{ seconds}$,then:

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